If an equation f(x) = c has a root x=a of multiplicity 2 then f'(a) = 0 (I know this is the pre-calculus forum, but I don't think you'll be able to do this question without using differentiation). If the root has multiplicity greater than 2, then also f"(a) = 0.

For the equation x(x+1)(x+2)...(x+2009) = c, take logs to get . Then let . Differentiate to get . Then differentiate again to get . Since all the terms in that last sum have the same sign, it's obvious that their sum cannot be zero. So has no roots of multiplicity greater than 2.

To find the number of roots of multiplicity 2, notice that the derivative of a function has a root between each pair of consecutive roots of the function. Since x(x+1)(x+2)...(x+2009) = 0 has 2010 roots, its derivative will have 2009 roots.